Question

Two stationary sources each emit waves of wavelength $\lambda$. An observer moves from one source to another with velocity $u$. Then, the number of beats per second heard by the observer is:

[Assume, medium is stationary ; the source and the observer are collinear]

• A. $\frac{2u}{\lambda }$
• B. $\frac{u}{\lambda }$
• C. $\sqrt{u\lambda }$
• D. $\frac{u}{2\lambda }$

Answer 1

The correct option is A $\frac{2u}{\lambda }$

Let both the sources have the same frequency $f_0.$

Frequency of each stationary source $f_0=\frac{v}{\lambda }$.
where $v$ is the velocity of wave in stationary medium.

Let us consider, the observer is moving from source$-1$ to source$-2$.

Then, from the definition of Doppler effect,

When the observer is moving away from the source,

$f_1=\frac{v-v_o}{v}\times f_0=\frac{1}{\lambda }(v-v_o)$

When the observer is moving towards the source,

$f_2=\frac{v+v_o}{v}\times f_0=\frac{1}{\lambda }(v+v_o)$

From the data given in the question, $v_o=u$

$\therefore$ Number of beats per second is given by
$|f_2-f_1|=[\frac{(v+u)}{\lambda }-\frac{(v-u)}{\lambda }]=\frac{2u}{\lambda }$